if 5tan⁡ϕ=tan⁡θ then the maximum value of 5tan2⁡(θ−ϕ) is

if $5 \tan ϕ = \tan θ$ then the maximum value of $5 \tan^{2} (θ - ϕ)$ is

A
1
B
4
C
6
D
9

Solution:

$t = \tan ϕ, \tan (θ - ϕ) = \frac{5 t - t}{1 + 5 t^{2}} = \frac{4 t}{1 + 5 t^{2}}$

5 $\tan^{2} (θ - ϕ) = \frac{80 t^{2}}{{(1 + 5 t^{2})}^{2}} = \frac{80}{{(\frac{1}{t} + 5 t)}^{2}} \leq \frac{80}{(2 \sqrt{5})^{2}} = 4$

since $A . M$ $\geq$ $G . M .$

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